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2 years ago

1. The choice of materials for an exciting playground slide should __________.

Physics

2 answers:

gavmur [86]2 years ago

4 0

have a large coefficient of kinetic friction between cloth and the slide material
40(0.3+0.5)=40(0.8)=32N
F-applied=Wbook/μk
When the tire is locked, the frictional force is greater but the car cannot be steered

Ugo [173]2 years ago

3 0

Answer:

1 have a small coefficient of kinetic friction between cloth and the slide material

2 12 N

3 F-applied=Wbook/μs

4 The static friction force is larger than the kinetic friction force

Explanation:

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A resistor and a capacitor are connected in series across an ideal battery having a constant voltage across its terminals. Long

tino4ka555 [31]

Answer:

A) equal to the battery's terminal voltage.

Explanation:

When the capacitor is fully charged after long hours of charging , its potential becomes equal to the emf of the battery and its polarity is opposite to that of battery . Hence net emf becomes equal . The capacitor itself becomes a battery which is connected in the circuit with opposite polarity . This results in the net emf and current becoming zero . There is no charging current when the capacitor is fully charged .

4 0

3 years ago

To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has

Roman55 [17]

Answer:

The magnitude of the acceleration is $a = 0.33 m/s^2$

The direction is $- \r k$ i.e the negative direction of the z-axis

Explanation:

From the question we are that

The mass of the particle $m = 1.8*10^{-3} kg$

The charge on the particle is $q = 1.22*10^{-8}C$

The velocity is $\= v = (3.0*10^4 m/s ) j$

The the magnetic field is $\= B = (1.63T)\r i + (0.980T) \r j$

The charge experienced a force which is mathematically represented as

$F = q (\= v * \= B)$

Substituting value

$F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \ ( 1.63 \r i + 0.980 \r j )T)$

$= 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \ X \ \ \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \ \r j))$

$= (-5.966*10^4 N) \r k$

Note :

$i \ \ X \ \ j = k \\\\j \ \ X \ \ k = i\\\\k \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \ X \ \ j = -i\\\\i \ \ X \ \ k = - j\\$

Now force is also mathematically represented as

$F = ma$

Making a the subject

$a = \frac{F}{m}$

Substituting values

$a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}$

$= (-0.33m/s^2)\r k$

$= 0.33m/s^2 * (- \r k)$

6 0

3 years ago

A 75-g projectile traveling at 600 m /s strikes and becomes embedded in the 40-kg block, which is ini-tially stationary. Compute

levacccp [35]

Explanation:

The given data is as follows.

Mass, m = 75 g

Velocity, v = 600 m/s

As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.

$m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

where, $m_{1}$ = mass of the projectile

$m_{2}$ = mass of block

v = velocity after the impact

Now, putting the given values into the above formula as follows.

$m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

$75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v$

= $\frac{45}{50.075}$

v = 0.898 m/s

Now, equation for energy is as follows.

E = $\frac{1}{2}mv^{2}$

= $\frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}$

= 13500 J

Now, energy after the impact will be as follows.

E' = $\frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}$

= 20.19 J

Therefore, energy lost will be calculated as follows.

$\Delta E$ = E E'

= (13500 - 20) J

= 13480 J

And, n = $\frac{\Delta E}{E}$

= $\frac{13480}{13500} \times 100$

= 99.85

= 99.9%

Thus, we can conclude that percentage n of the original system energy E is 99.9%.

7 0

3 years ago

Read 2 more answers

NEED ANSWER ASAP!!!!!!

olga55 [171]

Answer:

yes. why do you need this answered asap? lol

5 0

3 years ago

Read 2 more answers

Can someone explain it with steps?

Anton [14]

Answer:

Option C is the correct answer

Explanation:

Distance travelled by car during reaction time

$=15\times0.4\\\\=6m$

The car stopped before hitting the animal by $1 m$

Distance travelled during deceleration is $21-6-1=14m$

Hence by $v^2=u^2+2as$

We have

$0^2=15^2+2 \cdot a \cdot 14\\\\a=\frac{-225}{28} \\\\=-8.03m/s^2$

Option C is the correct answer

5 0

3 years ago

Read 2 more answers